on the right side, but all equal in value on one side. The pre-
liminary polygon for Ox pole is not required, for the line of
resultant for the links 0 3 and O 6 is evidently passing through
point a. Join this point to A and B and find the pole O2 fo*r a
polygon passing through A and B. Next join A B and produce
leftward. This is the line of rotation. Produce line d e, the
link O 9 till it meets A B produced in ^ ; and join C e^ giving the
link OQ in the final polygon. Draw a line in force diagram || C ^
from point 9, and another line O2 0F || A B. The intersection of
these lines is Or the pole of the final polygon, which may now be
completed in the space diagram, and which will, of course, pass
through the three points A, B, and c. This is the line of pressures,
and coincides very nearly with the arc of the bridge from c to A,
It deviates appreciably at the abutments, and causes a bending
mount on the left of O 12 x hj, but a larger moment of O o x/Y on
the right. The thrust between the abutments is nearly horizontal
and is found by joining k m in the space diagram. The parallel
p OF in the force diagram gives the value.
P. 474. Velocities:
Example 75. A motor-car moves in a horizontal circle of 300 ft.
radius. The track slopes sideways at an angle of 10° with the hori-
zontal A plumb-line on the car makes an angle of 12° with a
r of Molor-Car. ' Eig. 968.